Answer:
the lens you must select has an angle of 143º measured with respect to the horizontal, this angle is 53º with respect to the vertical.
Explanation:
The glare is caused by the reflection of light in the water, the polarization of the reflected light is polarized in a direction parallel to the surface of the water, the polarization is total for the angles
n = tan
\theta_{p} = tan⁻¹ n
the refractive index for seawater is 1.33
\theta_{p}= tan⁻¹ 1.33
\theta_{p} = 53º
for this angle the light is totally polarized, for the other angles the polarization is partial.
Based on this, the lenses must eliminate this polarization, so its polarization direction must have 90º with respect to this polarization,
\theta_{lens} = 53 +90
\theta_{lens}= 143º
Therefore, the lens you must select has an angle of 143º measured with respect to the horizontal, this angle is 53º with respect to the vertical.
A lens that could work is one that is polarized 45º with respect to the vertical.
B identifying the types of data to be gathered
Answer and Explanation:
The answer is attached below
Answer:
<h2>35</h2>
Explanation:
According to snell's law which states that the ratio of the sin of incidence (i) to the angle of refraction(n) is a constant for a given pair of media.
sini/sinr = n
n is the constant = refractive index
Since the diver shines light up to the surface of a flat glass-bottomed boat, the refractive index n = nw/ng
nw is the refractive index of water and ng is that of glass
sini/sinr = nw/ng
given i = 30°, nw = 1.33, ng = 1.5, r = angle the light leave the glass
On substitution;
sin 30/sinr = 1.33/1.5
1.5sin30 = 1.33sinr
sinr = 1.5sin30/1.33
sinr = 0.75/1.33
sinr = 0.5639
r = arcsin0.5639
r ≈35°
angle the light leave the glass is 35°
Answer:
Please selection one of the following